3.524 \(\int \frac{(a+b x)^{3/2}}{\sqrt{x}} \, dx\)

Optimal. Leaf size=71 \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{4 \sqrt{b}}+\frac{3}{4} a \sqrt{x} \sqrt{a+b x}+\frac{1}{2} \sqrt{x} (a+b x)^{3/2} \]

[Out]

(3*a*Sqrt[x]*Sqrt[a + b*x])/4 + (Sqrt[x]*(a + b*x)^(3/2))/2 + (3*a^2*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])
/(4*Sqrt[b])

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Rubi [A]  time = 0.0215816, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {50, 63, 217, 206} \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{4 \sqrt{b}}+\frac{3}{4} a \sqrt{x} \sqrt{a+b x}+\frac{1}{2} \sqrt{x} (a+b x)^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)/Sqrt[x],x]

[Out]

(3*a*Sqrt[x]*Sqrt[a + b*x])/4 + (Sqrt[x]*(a + b*x)^(3/2))/2 + (3*a^2*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])
/(4*Sqrt[b])

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(a+b x)^{3/2}}{\sqrt{x}} \, dx &=\frac{1}{2} \sqrt{x} (a+b x)^{3/2}+\frac{1}{4} (3 a) \int \frac{\sqrt{a+b x}}{\sqrt{x}} \, dx\\ &=\frac{3}{4} a \sqrt{x} \sqrt{a+b x}+\frac{1}{2} \sqrt{x} (a+b x)^{3/2}+\frac{1}{8} \left (3 a^2\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx\\ &=\frac{3}{4} a \sqrt{x} \sqrt{a+b x}+\frac{1}{2} \sqrt{x} (a+b x)^{3/2}+\frac{1}{4} \left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )\\ &=\frac{3}{4} a \sqrt{x} \sqrt{a+b x}+\frac{1}{2} \sqrt{x} (a+b x)^{3/2}+\frac{1}{4} \left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a+b x}}\right )\\ &=\frac{3}{4} a \sqrt{x} \sqrt{a+b x}+\frac{1}{2} \sqrt{x} (a+b x)^{3/2}+\frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{4 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.0972369, size = 69, normalized size = 0.97 \[ \frac{1}{4} \sqrt{a+b x} \left (\frac{3 a^{3/2} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{b} \sqrt{\frac{b x}{a}+1}}+\sqrt{x} (5 a+2 b x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)/Sqrt[x],x]

[Out]

(Sqrt[a + b*x]*(Sqrt[x]*(5*a + 2*b*x) + (3*a^(3/2)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(Sqrt[b]*Sqrt[1 + (b*x)
/a])))/4

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Maple [A]  time = 0.003, size = 78, normalized size = 1.1 \begin{align*}{\frac{1}{2} \left ( bx+a \right ) ^{{\frac{3}{2}}}\sqrt{x}}+{\frac{3\,a}{4}\sqrt{x}\sqrt{bx+a}}+{\frac{3\,{a}^{2}}{8}\sqrt{x \left ( bx+a \right ) }\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ){\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)/x^(1/2),x)

[Out]

1/2*(b*x+a)^(3/2)*x^(1/2)+3/4*a*x^(1/2)*(b*x+a)^(1/2)+3/8*a^2*(x*(b*x+a))^(1/2)/(b*x+a)^(1/2)/x^(1/2)*ln((1/2*
a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))/b^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.49584, size = 311, normalized size = 4.38 \begin{align*} \left [\frac{3 \, a^{2} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (2 \, b^{2} x + 5 \, a b\right )} \sqrt{b x + a} \sqrt{x}}{8 \, b}, -\frac{3 \, a^{2} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (2 \, b^{2} x + 5 \, a b\right )} \sqrt{b x + a} \sqrt{x}}{4 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^(1/2),x, algorithm="fricas")

[Out]

[1/8*(3*a^2*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(2*b^2*x + 5*a*b)*sqrt(b*x + a)*sqrt(
x))/b, -1/4*(3*a^2*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (2*b^2*x + 5*a*b)*sqrt(b*x + a)*sqrt(
x))/b]

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Sympy [A]  time = 3.62097, size = 75, normalized size = 1.06 \begin{align*} \frac{5 a^{\frac{3}{2}} \sqrt{x} \sqrt{1 + \frac{b x}{a}}}{4} + \frac{\sqrt{a} b x^{\frac{3}{2}} \sqrt{1 + \frac{b x}{a}}}{2} + \frac{3 a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{4 \sqrt{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)/x**(1/2),x)

[Out]

5*a**(3/2)*sqrt(x)*sqrt(1 + b*x/a)/4 + sqrt(a)*b*x**(3/2)*sqrt(1 + b*x/a)/2 + 3*a**2*asinh(sqrt(b)*sqrt(x)/sqr
t(a))/(4*sqrt(b))

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^(1/2),x, algorithm="giac")

[Out]

Timed out